If [tex]0<A<B[/tex] then(adsbygoogle = window.adsbygoogle || []).push({});

[tex]

B < (A^n + B^n)^{1/n} < 2^{1/n} B,\quad\quad n=1,2,3,\ldots

[/tex]

and by using knowledge [tex]2^{1/n}\to 1[/tex] as [tex]n\to\infty[/tex], we get

[tex]

\lim_{n\to\infty} (A^n + B^n)^{1/n} = B.

[/tex]

Now, I'm interested to know what can we say if [tex]-B<A<0<B[/tex]. With odd n the inequality

[tex]

B\leq (A^n + B^n)^{1/n}

[/tex]

is wrong, so there does not seem to be any reason to believe that the limit would still be B. But is the limit still B?

For example, set A=-1, B=2. The numbers

n, (-1)^n + 2^n, ((-1)^n + 2^n)^(1/n)

turn out as follows.

1, 1, 1

2, 5, 2.2361...

3, 7, 1.9129...

4, 17, 2.0305...

5, 31, 1.9873...

6, 65, 2.0052...

7, 127, 1.9978...

The 2^n appears to be dominating the limit quite well. Is the limit of this 2 really? What about more general conditions for A and B?

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# (A^n+B^n)^(1/n) limit stuff

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